Problem: Let $h(x)=\dfrac{x^2+5}{x^2-1}$. Find $h'(0)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-5$ (Choice B) B $1$ (Choice C) C $-12$ (Choice D) D $0$
Explanation: Let's first find the derivative of $h$, i.e. the expression for $h'(x)$. Then, we can plug $x=0$, into $h'(x)$ and evaluate. $h$ is a rational function. To find the derivative of rational functions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = h ′ ( x ) = d d x ( x 2 + 5 x 2 − 1 ) = ( x 2 − 1 ) d d x ( x 2 + 5 ) − ( x 2 + 5 ) d d x ( x 2 − 1 ) ( x 2 − 1 ) 2 = ( x 2 − 1 ) ( 2 x ) − ( x 2 + 5 ) ( 2 x ) ( x 2 − 1 ) 2 = 2 x 3 − 2 x − 2 x 3 − 10 x ( x 2 − 1 ) 2 = − 12 x ( x 2 − 1 ) 2 The quotient rule Differentiate ( x 2 + 5 ) & ( x 2 − 1 ) Expand \begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac{x^2+5}{x^2-1}\right) \\\\ &=\dfrac{(x^2-1)\dfrac{d}{dx}(x^2+5)-(x^2+5)\dfrac{d}{dx}(x^2-1)}{(x^2-1)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{(x^2-1)(2x)-(x^2+5)(2x)}{(x^2-1)^2}&&\gray{\text{Differentiate }(x^2+5)\text{ & }(x^2-1)} \\\\ &=\dfrac{2x^3-2x-2x^3-10x}{(x^2-1)^2}&&\gray{\text{Expand}} \\\\ &=\dfrac{-12x}{(x^2-1)^2} \end{aligned} So we found that $h'(x)=\dfrac{-12x}{(x^2-1)^2}$. Now let's plug $x= {0}$ into $h'$ : $\begin{aligned} &\phantom{=}h'( {0}) \\\\ &=\dfrac{-12( {0})}{(( {0})^2-1)^2} \\\\ &=\dfrac{0}{(-1)^2} \\\\ &=0 \end{aligned}$ In conclusion, $h'(0)=0$.